Problems & Puzzles: Puzzles

Puzzle 7.- Palindrome - Primes: 2 questions:

1. Patrick De Geest, observes that:

10499 + 10501 + 10513 = 31513

is the first case of a prime palindrome (31513) obtained adding three consecutive primes, the central one of them being a palindrome prime also (10501), and asks if this will ever happen again?

( http://www.worldofnumbers.com/palprim5.htm )

2. By my own I ask if exists any palindrome prime obtained adding the first k primes, that is to say:

What is the first k, such that:

Nk = 2 + 3 + 5 + 7 + 11 +…..+ pk = Prime Palindrome?

Solution

Question b.

Jud McCranie sends his results on this problem : 

"One of your puzzles asks for the sum of the first k primes to be a palindromic prime. Except for the trivial solutions k=1, sum=2 and k=2, sum=5; there aren't any more for k <= 54,400,028 (Pk = 1073741989, sum Nk = 28422918403819825)" (9/July/98) 

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Vasiliy Danilov sent us (Fri. 17 Jul. 1998) the following palindromes (not-primes) of his search for this puzzle : 

k             Pk            Sum k 
8              19              77 
7693       78347       285080582 
8510       87641       352888253 
12941     139241     854848458 
146134   1959253  137372273731 
637571   9564097  2939156519392 

Pk < 100000000 and program stopped. 

He also sent us some references at the Neil Sloane’s records (A038582 , A038584, A038583,) partially related with this puzzle. 
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Judson McCranie adds this two lines (Jul/23/98) of palindromic not-primes, as a result of his search ... "I've now checked k through 105097565"

    k            Pk         Sum k (palindromic)

 27198825      516916921    6833383883833386
 53205635     1048924213   27155268786255172
 
So, it seems to me that the answer to this puzzle is "negative"...

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Question a.

Almost five years later, Giovanni Resta came up with several (8) solutions, as the asked by Patrick

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One week later J. K. Andersen came up with the following big solutions to question a), gotten with a clever approach (as usual):

Question a

I have found many big solutions with a pattern.

Let p be a palindrome where the first/last digit and all digits in even positions are in {0,1,2} and all other digits (odd internal positions) are in {7,8,9}. Then 3p+20 is also a palindrome. Some of Resta's solutions follow this pattern. My strategy: Generate such palindromes p and 3p+20. If both are prp then find the prp right before and after p. If they sum to 3p+20 then we have a probable solution and prove the 4 primes.

First a C program using Michael Scott's Miracl bigint library found the solutions with prp's. PrimeForm/GW double-checked that the 3 prp's are consecutive.

A 73-digit solution with middle palindromic prime:

p = 1070707070707070707070707071829182829282819281707070707070707070707070701

The solution equation is (p-52) + (p) + (p+72) = (3p+20)

= 3212121212121212121212121215487548487848457845121212121212121212121212123

The 4 primes were proved with Tony Forbes' VFYPR.

A 303-digit solution:

p = 10^302 + 10*(7*10^301-70)/99 + 10^140*22001012200100221010022 + 1 Let w be "0" followed by 69 concatenations of "70". Then p is the concatenation 1w92708082907170928080729w1 Solution equation: (p-138) + (p) + (p+158) = (3p+20) These 4 primes were proved with Marcel Martin's Primo.

After trying 25098 prp's as the middle palindrome I got this 527-digit

solution:

p = 10^526 + 10*(7*10^525-70)/99 + 10^251*2102210022021202200122012 + 1 Let w be "0" followed by 124 concatenations of "70". Then p is the concatenation 1w728092807290919092708290827w1 The equation is (p-180) + (p) + (p+200) = (3p+20), proved with Primo.

The complexity of my algorithm is around O(digits^5). I will not search a titanic solution. It is about 25 times harder than 527 digits and would probably take weeks.

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